Answer the following questions: I. Are the enthalpies of atomisation of Zinc and Copper matched correctly? Justify your answer. Element Enthalpy of atomisation/ kJmol^-1 --- --- Zinc 339 Copper 130 II. Out of sulphuric acid and hydrochloric acid, which acid will you prefer for permanganate titrations and why? III. 5NO_2^- + 2MnO_4^- + 6H^+ →

ChemistryCBSE Class 12 · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2025-26: question 28

From the official CBSE sample question paper for 2025-26, section C, question 28, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28 · 3 marks · Short answerThe d- and f-Block Elements

Answer the following questions:
I. Are the enthalpies of atomisation of Zinc and Copper matched correctly? Justify your answer.

Element	Enthalpy of atomisation/ $kJmo l^{- 1}$
Zinc	339
Copper	130

II. Out of sulphuric acid and hydrochloric acid, which acid will you prefer for permanganate titrations and why?
III. $5N O_{2}^{-} + 2Mn O_{4}^{-} + 6 H^{+} \to$

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Enthalpy of atomisation depends on the strength of metallic bonding, which is linked to unpaired d electrons.

Hint 2

Which acid can itself be oxidised by permanganate?

Hint 3

In acidic medium, MnO₄⁻ is reduced to Mn²⁺ and nitrite is oxidised to nitrate.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

I. No, the values are interchanged. Zinc ([Ar]3d $^{10}$ 4s $^{2}$ ) has no unpaired d electrons, so its metallic bonding is weak and its enthalpy of atomisation is low, **130 kJ mol $^{- 1}$ . Copper has stronger metallic bonding, so its value is 339 kJ mol $^{- 1}$ **.
1 mark
II. Sulphuric acid is preferred. HCl is oxidised by KMnO $_{4}$ to Cl $_{2}$ , so part of the permanganate is used up and the titre is wrong. Dilute H $_{2}$ SO $_{4}$ is not oxidised by MnO $_{4}^{-}$ .
1 mark
III. $5 NO_{2}^{-} + 2 MnO_{4}^{-} + 6 H^{+} \to 2 Mn^{2 +} + 5 NO_{3}^{-} + 3 H_{2} O$
1 mark

Answer: I. No; Zn = 130 and Cu = 339 kJ/mol (Zn has no unpaired d electrons, so weaker metallic bonding). II. H₂SO₄, because HCl is oxidised to Cl₂ by KMnO₄. III. 2Mn²⁺ + 5NO₃⁻ + 3H₂O

Where the marks are

States 'No' and justifies using the absence of unpaired electrons in Zn (1 mark)
H₂SO₄ chosen, since HCl is oxidised to Cl₂ (1 mark)
Correct products of the balanced equation (1 mark)

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Chapter: The d- and f-Block Elements (notes, formulas, where marks are lost and every past question)

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CBSE Class 12 Chemistry sample paper 2025-26: question 28

Hints

Step-by-step solution

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